問題如下...
prove that : as n -> infinity, x^n -> 0 if | x | < 1.
given E > 0, find a number N(E) - (which depends on E),
for which m > N(E) => | x^m | < E.
請各位幫幫手...唔該~ ^ ^
關於convergent series 的問題... 急急急!!
2008-12-11 8:31 am
回答 (2)
2008-12-11 8:49 am
✔ 最佳答案
Since |x^n - 0 |=|x^n|
if |x^n|<ε
if x bigger then 1, then the ratio between two consecutive terms should be bigger than 1. So x^n will tend to infinity which is a contradiction. So | x | < 1
Now x^n<ε
<=>n ln x<lnε
<=>n>lnε/ln x
So if N(ε)=[lnε/ln x]+1, then for m > N(ε) => | x^m | < E.
2008-12-11 8:57 am
∀ε>0, ∃N= [lnε/ln x]+1 which is a natural number
such that
∀n>N, |x^n| < |x^N| < |x^(lnε/ln x)| =ε
Therefore lim(n→∞) |x^n|=0 for |x|<1
such that
∀n>N, |x^n| < |x^N| < |x^(lnε/ln x)| =ε
Therefore lim(n→∞) |x^n|=0 for |x|<1
參考: ME
收錄日期: 2021-04-25 16:59:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081211000051KK00055